Question

定义在R上的奇函数f(x),且f(x)的最小正周期是π,当x∈[0,π/2]时

f(x)=sinx,则f(5π/3)=

Answer 1

答：
定义域在R上奇函数f(x)满足：
f(-x)=-f(x)，f(0)=0
f(x+π)=f(x)
0<=x<=π/2时，f(x)=sinx
f(5π/3)=f(2π/3+π)=f(2π/3)=f(-π/3+π)=f(-π/3)=-f(π/3)=-sin(π/3)=-√3/2
所以：
f(5π/3)=-√3/2

Answer 2

因为f(x)是周期函数，5π/3减去2π,函数值不变，f(5π/3)=(-π/3)，又因为是奇函数，f(-π/3)=-f(π/3)=-1/2